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Zina [86]
3 years ago
5

Solve using any method.

Mathematics
2 answers:
Klio2033 [76]3 years ago
7 0
The answer should be 4. (2, 1)
GaryK [48]3 years ago
3 0
The fourth one since -2(2) would be -4 then the y is 1 which would make it into -4+1=-3 and it’s the same as the fourth one
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A football coach walks 12 meters westward, then 16 meters eastward,
tino4ka555 [31]

Answer:

Total distance traveled = 56 m

Total displacement is 12 m

Direction of coach is westward

Step-by-step explanation:

It is given that football coach first go 12 meter westward then 16 meter eastward.

So no position of coach in eastward direction = 16 - 12 = 4 m

And then 22 m west westward

Therefore position of coach now is 22-4 = 18 m westward

And finally 6 m eastward

Therefore final position of coach is 18-6= 12 m westward

So displacement of coach is 12 m and direction is westward.

Total distance traveled = 12+16+22+6= 56 m

4 0
2 years ago
Read 2 more answers
A new club sent out 172 coupons to boost sales for next year's memberships. They provided 3 times as many to potential members
cricket20 [7]

Answer: 172: 4= 43

43× 3= 129

172-129= 43

Odp.Wysłali 43 kupony do istniejących członków.

Step-by-step explanation:1.bo wysłali 3×wiecej i jeszcze są istniejący członkowie

2.ile wyslali potencialnym

3.i ile wyslali do istniejacych

4 0
3 years ago
koda Buys 0.75 kg of cortos,which ia 5times the mas of the union he also Buys. How much does the union weigh?
kari74 [83]

Answer:

0.15kg

Step-by-step explanation:

Given data

We are told that

0.75 kg of cortos weights 5times the mas of the onion

We want to find the mass of 1 onion

Hence

0.75 kgcortos = 5 onions

      x cortos = 1 onions

x= 0.75/5

x= 0.15kg

Hence 1 onion will weigh 0.15kg

4 0
2 years ago
Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
2 years ago
12. If 5t + 2 = 6, then t =
tankabanditka [31]
It would be t=0.8, mainly bc you subtract two from six , that reduces it to four 
then you divide four by five and it should give you your answer
and to chack your work put .8 where t was and solve 
6 0
3 years ago
Read 2 more answers
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